Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Given the equation 2(x^4) + 4(x^2) + 1 = 0 find all real solutions....?
Intuitively, I can see for any x the value of the function will be positive, and at x = 0 the y value is at its minimum, 1. So there are no real solutions, how can I prove this? I tried substitution (let w = x^2) but the discriminant I keep finding to be positive, which I was hoping to find negative.
2 Answers
- Jeff AaronLv 77 years agoFavorite Answer
Let w = x^2, so we have:
2w^2 + 4w + 1 = 0
w = (-4 +/- sqrt(4^2 - 4*2*1)) / (2*2)
w = (-4 +/- sqrt(16 - 8)) / 4
w = (-4 +/- sqrt(8)) / 4
w = -1 +/- sqrt(0.5)
w =~ -0.29289321881345247559915563789515 or -1.7071067811865475244008443621048
x^2 =~ -0.29289321881345247559915563789515 or -1.7071067811865475244008443621048
Since x^2 is negative, then x can't be real.
- daniel mLv 67 years ago
let y = x^2
then y^2 = x^4
2y^2 + 4y + 1 = 0
quad formula
y = (-4 +/- sqrt(8))/4
x^2 = (-4 +/- sqrt(8))/4
x = +/- sqrt((-4 +/- sqrt(8)))/2
-4 - sqrt(8) will be negative, obviously...
4 > sqrt(8), so -4 + sqrt(8) is also negative
you can't sqrt negative numbers and get real solutions... so there are no real solutions to this
